Thus A and B being the given points, the curve passing through Fis a hyperbola. If another similar curve pass through E, so that AE = BF, these branches are called opposite hyperbolas.

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2. The given points are named the foci; that part of the line that joins them, intercepted by opposite hyperbolas, the transverse axis;

and the middle of the same line, the centre.

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3. The conjugate axis is a straight line, passing through the centre, perpendicular to the transverse, and limited by a circle described from one extremity of the transverse with the distance of either focus from the centre as a radius.

4. If two other hyperbolas be described, having the conjugate for their transverse axis, and their foci at the same distance from the centre as the foci of the former, these are also called opposite hyperbolas, and have the transverse of the former for their conjugate axis. These are the branches through G and H.

5. A diameter is a straight line drawn through the centre, and terminated by opposite hyperbolas.

6. An ordinate to a diameter is a straight line, terminated by the hyperbola, and bisected by that diameter produced.

7. An external ordinate to a diameter is a straight line, terminated by opposite hyperbolas, and bisected by that diameter, or the same produced.

8. Two diameters are said to be conjugate to one another when they are mutually parallel to each other's ordinates. 9. A third proportional, to any diameter, and its conjugate, is called its parameter.

10. The eccentricity is the distance between the centre and either focus.

11. An assymptote of the hyperbola is a straight line, which, being produced indefinitely, does not meet, but con

tinually approaches the curve, so as to come within less than any given distance from it.

12. When the assymptotes are perpendicular, the hyperbola is said to be rectangular.

PROPOSITION I.

If from a point in any hyperbola, straight lines be drawn to the foci, their difference is equal to the transverse axis. Let DF, ET, be opposite hyperbolas, of which A, B, are the foci, C the centre, EF the transverse axis, and D any point in one of

the curves; then AD-DB-EF.

For AD-DB =AF—FB (III. Def. 1) AE+

=

EF-FB-EF.

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COR. 1.-If two straight lines be drawn to the foci, from a point without the opposite hyperbolas, their difference is less than the transverse axis, but, if from any point within either of them, it is greater.

Let M be a point within the hyperbola DF, and Na point without, between the curve and its conjugate axis. The lines AM, NB, necessarily meet the curve; let them meet it in the points D, D; then, since MD+DB>MB, AM MB is greater than AM (MD + DB), or EF, but since AD + DN AN, AN NB is less than (AD+DN) - NB, or EF.

COR. 2.-A point is within, in, or without, the curve, according as the difference of its distances from the foci is greater, equal, or less, than the transverse axis. COR. 3.-The conjugate axis is bisected in the centre. For (see figure to Def. 1) EG ACEH, and hence angle GH, and those at C are right angles (III. Def. 3) ; therefore (Pl. Ge. I. 26) GC = CH.

COR. 4. The rectangle under the segments, into which the transverse is divided in one of the foci, is equal to the square of the semi-conjugate axis. For CG2 GF2 CF2 - CB2 — CF2 — EB · BF.

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COR. 5.-The square of the distance of the foci is equal to the sum of the squares of the transverse and conjugate axis.

COR. 6.-A perpendicular to the transverse, at its extremity, is a tangent to the hyperbola.

For (Pl. Ge. II. c. Cor. 1) (AQ+QB) (AQ—QB)= (AF+FB) (AF-FB) = AB EF. But AQ+QB AB, therefore AQ-QB EF, and the point Q is without the curve (III. 1, Cor. 2).

PROPOSITION II.

The straight line which bisects the angle, contained by two straight lines, drawn from any point in the hyperbola to the foci, is a tangent to the curve in that point.

Let D be any point in the hyperbola DFL, from which AD, DB, are drawn to the foci, and let the angle BDI be bisected by the line

DT, then is DT a tangent to the curve in the point D.

For, let any other point R be assumed in DT, and DI being made equal to DB, let BI, RB, RI, be drawn, and suppose A, R, joined. The line DNT, which bisects the ver

tical angle of the isosceles triangle BDI, also bisects the base BI at right angles (Pl. Ge. I. 4). Hence RB = RI, and AR- RB: ARRI, and therefore less than AI or EF. Consequently, the point R is without the hyperbola.

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COR. 1.-The method of drawing a tangent from a given point in the curve, also of drawing a tangent parallel to a line given in position not parallel to the transverse, is evident.

To draw a tangent parallel to a given line; from B draw BI perpendicular to the given line, and make AI = EF, then a line, as NR, bisecting BI perpendicularly, is a tangent.

COR. 2.-There cannot be more than one tangent to the hyperbola at the same point.

For the difference between the lines AD, DB, that make equal angles with DN, is less than that of any other two lines from A and B to a point in DN. The rest of the proof is similar to that of Cor. 3, Prop. 2, on the ellipse. COR. 3.-Every tangent bisects the angle contained by straight lines drawn to the foci from the point of contact; or, which is the same, these lines make equal angles with the tangent.

COR. 4.-Every tangent to the same hyperbola meets the transverse between its vertex and the centre.

For (Pl. Ge. VI. 3) AD: DB = AT: TB; but AD is always greater than DB by EF, therefore AT is always greater than TB.

COR. 5.-A straight line drawn from the centre to meet a tangent, and parallel to the line joining the point of contact, and one of the foci, is equal to half the transverse axis.

Since INNB, and AC = CB, CN is parallel to AI, and equal to the half of it, or of EF.

COR. 6. A perpendicular to a tangent, from one of the foci, and a parallel to the line joining the other focus, and the point of contact from the centre, meet the tangent in the same point.

COR. 7.-If a chord pass through one of the foci, and the tangents at its extremities be produced to meet, the straight line that joins the point of concourse and the focus, will be perpendicular to the cord.

From O draw a line perpendicular to AD, and if it does not meet it in I, let it meet it in a point I'. Then (Pl. Ge. VI. н) AL+ DI′ = semi-perimeter of the triangle ADL. But since AD DB - EF AL-LB, therefore AL+ DB AD+LB; and hence AL+ DB semi-perimeter,

and therefore DI' = DB, and the point I' must therefore coincide with I. Therefore in the triangles ODI, ODB, the sides OD, DI, are equal to OD, DB, and the contained angles at D are equal; therefore angle OBD = OID = a right angle; and OB is perpendicular to DL.

PROPOSITION III.

From any point in an hyperbola, a perpendicular being let fall upon the transverse, and straight lines drawn to the foci; as half the transverse is to the distance of the centre from the focus, so is the distance of the centre from the perpendicular to half the sum of the straight lines drawn to the foci.

Let D be the point in the curve, from which DA, DB, are drawn to the foci, and DK perpendicular to EF; then CF: CBCK: (AD + DB) (figure Prop. 1).

For, let EFL AD, then LFDB, and 2CL= AD + DB. But (AD + DB) · (AD — DB) = AB · 2CK (Pl. Ge. II. c. Cor. 1), or EF 2CL AB 2CK; therefore EF: AB2CK: 2CL, and CF : CB = CK: CL.

PROPOSITION IV.

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Any two chords which intersect each other in the focus of an hyperbola, are to one another directly as the rectangles under their segments.

Let the chords DO, MI, intersect one another in the focus B of the hyperbola DFO, DO: MI = DB · B0 : MB BI.

For, let the tangents at D, O, meet each other in R, through C let NQ be drawn parallel to DO, meeting the tangents in N, Q, and let BR meet NQ in P. It has been proved that CN or CQ = CF (III. 2, Cor. 5), therefore NQ=EF; that CF: CB CK : CL (III.3), therefore CF CL =CB CK, and that DBR is a right angle (III. 2, Cor. 7), therefore the triangles DBK, BCP, are similar, and

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M

FABK

DB CP BC BK. Hence if FL= DB, CF CF CL

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